Question: Find $\lim_{x\to\infty}\dfrac{1}{\sin(x)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $0$ (Choice C) C $1$ (Choice D) D The limit doesn't exist
Answer: When dealing with limits that include $\sin(x)$, it's important to remember that $\lim_{x\to\infty}\sin(x)$ doesn't exist, as $\sin(x)$ keeps oscillating between $-1$ and $1$ forever. ${2}$ ${4}$ ${6}$ ${8}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ ${2}$ $y$ $x$ $y=\sin(x)$ This doesn't necessarily mean that our limit doesn't exist. Think what happens to $\dfrac{1}{\sin(x)}$ as $x$ increases towards positive infinity. While $1$ remains constant forever, $\sin(x)$ oscillates from $-1$, to $0$, to $1$, to $0$, to $-1$ again. The result is a graph that goes up and down forever, with vertical asymptotes every now and then. ${50}$ ${100}$ ${150}$ ${\llap{-}50}$ ${\llap{-}100}$ ${\llap{-}150}$ ${50}$ ${\llap{-}50}$ $y$ $x$ $y=\dfrac{1}{\sin(x)}$ This limit doesn't approach any specific value as $x$ increases towards infinity. In conclusion, $\lim_{x\to\infty}\dfrac{1}{\sin(x)}$ doesn't exist.